Science · Chemistry ★★☆ Medium UNIT 3 OF 0

Chemical Bonding — Free Chemistry Review Games.

This unit covers ionic bonds, covalent bonds, Lewis structures and electronegativity — essential concepts for Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.

📋 60 questions ⏱ ~25 min
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All 60 questions below, each with the worked answer and a written explanation. Click any question to expand it.

Q1. What type of bond forms when electrons are transferred between atoms?
A Covalent
B Ionic
C Metallic
D Hydrogen

Ionic bonds form when one atom transfers electrons to another, creating oppositely charged ions that attract each other.

Q2. What type of bond involves the sharing of electrons?
A Ionic
B Covalent
C Metallic
D Nuclear

Covalent bonds form when two atoms share one or more pairs of electrons.

Q3. How many valence electrons does carbon have?
A 2
B 4
C 6
D 8

Carbon is in Group 14 and has 4 valence electrons, allowing it to form 4 covalent bonds.

Q4. What is the octet rule?
A Atoms need 8 protons to be stable
B Atoms tend to gain, lose, or share electrons to have 8 valence electrons
C All molecules have 8 atoms
D There are 8 types of bonds

The octet rule states that atoms tend to form bonds to achieve 8 electrons in their outer shell, like noble gases.

Q5. An ionic bond most commonly forms between which types of elements?
A Two metals
B Two nonmetals
C A metal and a nonmetal
D Two noble gases

Ionic bonds typically form between metals (which lose electrons) and nonmetals (which gain electrons).

Q6. What is a Lewis dot structure?
A A diagram of the nucleus
B A diagram showing valence electrons as dots around an element symbol
C A periodic table arrangement
D A type of chemical equation

Lewis dot structures represent an atom's valence electrons as dots around its chemical symbol to visualize bonding.

Q7. What is electronegativity's role in determining bond type?
A It has no role
B Large electronegativity differences create ionic bonds; small differences create covalent bonds
C It only affects metallic bonds
D Higher electronegativity always means ionic

A large electronegativity difference (>1.7) between atoms favors ionic bonding, while a small difference favors covalent bonding.

Q8. What is a polar covalent bond?
A A bond with equal electron sharing
B A bond with unequal electron sharing due to different electronegativities
C An ionic bond
D A bond between identical atoms

In a polar covalent bond, electrons are shared unequally because one atom has higher electronegativity, creating partial charges.

Q9. What type of bond holds atoms together in a metal?
A Ionic
B Covalent
C Metallic
D Van der Waals

Metallic bonds involve a 'sea' of delocalized electrons shared among metal atoms, giving metals conductivity and malleability.

Q10. What is a double bond?
A Two atoms bonded twice
B Two pairs of electrons shared between two atoms
C A bond twice as strong as ionic
D A metallic bond

A double bond involves the sharing of two pairs (four total) of electrons between two atoms.

Q11. Why does NaCl have a high melting point?
A It has covalent bonds
B Strong electrostatic forces between many Na+ and Cl- ions in the crystal lattice
C It is a gas at room temperature
D It has weak intermolecular forces

NaCl forms a crystal lattice with strong ionic bonds between many alternating Na+ and Cl- ions, requiring significant energy to break.

Q12. What is VSEPR theory used to predict?
A Reaction rates
B The three-dimensional shape of molecules
C Atomic mass
D Electron configuration

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular geometry by minimizing repulsion between electron pairs.

Q13. What shape does a molecule with 4 bonding pairs and no lone pairs have?
A Linear
B Trigonal planar
C Tetrahedral
D Bent

Four bonding pairs with no lone pairs arrange tetrahedrally (109.5 degree bond angles), as seen in methane (CH4).

Q14. Why is water (H2O) a polar molecule?
A It has ionic bonds
B Its bent shape creates an uneven distribution of charge
C Hydrogen is very electronegative
D It has no lone pairs

Water's bent shape (due to two lone pairs on oxygen) and the electronegativity difference between O and H create a net dipole moment.

Q15. What are intermolecular forces and how do they compare to intramolecular bonds?
A They are the same strength
B Intermolecular forces act between molecules and are weaker than bonds within molecules
C Intermolecular forces are stronger
D They only exist in gases

Intermolecular forces (like hydrogen bonds, dipole-dipole, London dispersion) act between molecules and are much weaker than covalent or ionic bonds within molecules.

Q16. What is a cation?
A An atom that has lost one or more electrons, giving it a positive charge
B An atom that has gained one or more electrons, giving it a negative charge
C A neutral atom with equal protons and electrons
D A molecule held together by covalent bonds

A cation forms when an atom loses electrons, leaving more protons than electrons and a net positive charge. The distractor "An atom that has gained one or more electrons, giving it a negative charge" describes an anion, not a cation. Recognizing that metals typically lose electrons to form cations is essential for predicting ionic compound formulas.

Q17. What is an anion?
A A negatively charged ion formed by gaining electrons
B A positively charged ion formed by losing electrons
C A neutral molecule with shared electrons
D An atom with a full outer shell before bonding

An anion carries a negative charge because it has gained one or more electrons, giving it more electrons than protons. The option "A positively charged ion formed by losing electrons" actually defines a cation, which is the opposite process. Nonmetals typically gain electrons to complete their octet, becoming anions.

Q18. How many valence electrons does oxygen have?
A 6
B 4
C 2
D 8

Oxygen is in Group 16 (VIA) of the periodic table, and elements in this group have 6 valence electrons in their outer shell. The choice "8" would represent a full octet, not oxygen's actual count before bonding. Knowing an element's group number lets you quickly determine its valence electron count for drawing Lewis structures.

Q19. Which of these best describes a nonpolar covalent bond?
A Electrons are shared roughly equally between two atoms
B Electrons are transferred completely from one atom to another
C Electrons are shared unequally, creating partial charges
D One atom pulls electrons away entirely, forming ions

A nonpolar covalent bond occurs when two atoms have very similar or identical electronegativities, so they share bonding electrons nearly equally. The distractor "Electrons are transferred completely from one atom to another" describes ionic bonding, a fundamentally different bonding type. Bonds between identical atoms, like \(\text{Cl}_2\), are classic examples of nonpolar covalent bonds.

Q20. What does a single line between two atoms in a Lewis structure represent?
A One shared pair of electrons (a single bond)
B Two shared pairs of electrons (a double bond)
C A lone pair of electrons on one atom
D An ionic bond between two ions

In Lewis structures, a single line drawn between two atoms represents one shared pair of electrons, forming a single covalent bond. The option "Two shared pairs of electrons (a double bond)" is incorrect because a double bond is shown with two lines, not one. Correctly counting lines versus dots is essential for accurately interpreting bonding and lone pairs in Lewis diagrams.

Q21. What are dots around an atom in a Lewis structure used to represent?
A Nonbonding (lone) pairs of valence electrons
B The number of protons in the nucleus
C The atomic mass of the element
D Electrons involved in metallic bonding

Dots placed around an atom's symbol in a Lewis structure represent valence electrons that are not involved in bonding, known as lone pairs. The distractor "The number of protons in the nucleus" confuses electron representation with a completely different atomic property. Correctly distinguishing bonding electrons (lines) from lone pairs (dots) is fundamental to drawing accurate Lewis structures.

Q22. Which pair of elements is most likely to form an ionic bond?
A Sodium and chlorine
B Carbon and hydrogen
C Nitrogen and oxygen
D Two chlorine atoms

Sodium is a metal with low electronegativity that readily loses an electron, while chlorine is a nonmetal with high electronegativity that readily gains one, producing a large electronegativity difference favorable for ionic bonding. The pairing "Carbon and hydrogen" involves two nonmetals with a small electronegativity difference, which instead forms a covalent bond. As a general rule, metal-nonmetal combinations tend to form ionic bonds due to their large electronegativity gap.

Q23. What term describes the tendency of an atom to attract shared electrons in a chemical bond?
A Electronegativity
B Ionization energy
C Atomic radius
D Electron affinity

Electronegativity specifically measures how strongly an atom attracts shared electrons within a covalent bond. "Ionization energy" instead measures the energy required to remove an electron from an isolated atom, which is a related but distinct concept. Electronegativity trends across the periodic table are the key tool for predicting bond polarity.

Q24. On the periodic table, electronegativity generally increases in which direction?
A Left to right across a period and bottom to top up a group
B Right to left across a period and top to bottom down a group
C Left to right across a period and top to bottom down a group
D Right to left across a period and bottom to top up a group

Electronegativity increases from left to right across a period because nuclear charge increases while shielding stays roughly constant, and it increases up a group because valence electrons are closer to the nucleus with less shielding. The option describing an increase "right to left across a period" reverses the correct periodic trend. Fluorine, located in the upper right of the periodic table, is the most electronegative element, which anchors this trend in memory.

Q25. Which of the following is a property typically associated with ionic compounds?
A High melting points and brittleness in solid form
B Low melting points and flexibility in solid form
C Good electrical conductivity in solid form
D Low solubility in water for most compounds

Ionic compounds form rigid crystal lattices held together by strong electrostatic forces, resulting in high melting points and brittleness when force is applied and disrupts the ion alignment. The claim of "Good electrical conductivity in solid form" is false because ions are fixed in place in the solid lattice and cannot move to carry charge until melted or dissolved. This solid-versus-molten conductivity distinction is a hallmark property used to identify ionic compounds experimentally.

Q26. What is a triple bond?
A Three shared pairs of electrons between two atoms
B One shared pair of electrons between three atoms
C Three separate ionic bonds in a compound
D A bond formed by transferring three electrons

A triple bond consists of three pairs of electrons, or six electrons total, shared between the same two atoms, as seen in \(\text{N}_2\). The distractor "One shared pair of electrons between three atoms" incorrectly describes electron sharing among three separate atoms rather than a stronger bond between two. Triple bonds are shorter and stronger than single or double bonds because more electron density holds the nuclei together.

Q27. What charge does a magnesium ion typically have after bonding?
A \(2+\)
B \(1+\)
C \(2-\)
D \(1-\)

Magnesium is in Group 2 and loses its two valence electrons to achieve a stable noble gas electron configuration, resulting in a \(2+\) charge. The option "\(2-\)" would require magnesium to gain two electrons, which is inconsistent with its low electronegativity and metallic character. Group number on the periodic table is a reliable predictor of the typical ionic charge for main-group metals.

Q28. Which type of bond forms between two atoms with an electronegativity difference greater than approximately \(1.7\)?
A Ionic bond
B Nonpolar covalent bond
C Metallic bond
D Hydrogen bond

An electronegativity difference above roughly \(1.7\) indicates that one atom pulls the bonding electrons so strongly that essentially complete electron transfer occurs, classifying the bond as ionic. A "Nonpolar covalent bond" instead requires a very small electronegativity difference, typically less than about \(0.4\). This numeric threshold is a useful rule of thumb for classifying bond type based on electronegativity difference, though it is a continuum rather than a sharp cutoff.

Q29. Using electronegativity values (\(\text{H} = 2.1\), \(\text{Cl} = 3.0\)), how would the H-Cl bond in \(\text{HCl}\) be classified?
A Polar covalent
B Nonpolar covalent
C Ionic
D Metallic

The electronegativity difference between H and Cl is about \(0.9\), which falls in the intermediate range that produces unequal electron sharing and partial charges, defining a polar covalent bond. The label "Ionic" requires a much larger difference, typically greater than \(1.7\), which this pair does not reach. Calculating electronegativity differences is a practical method for classifying real bonds rather than relying on general element type alone.

Q30. In the Lewis structure of \(\text{CO}_2\), what type of bonds connect the carbon atom to each oxygen atom?
A Double bonds
B Single bonds
C Triple bonds
D Ionic bonds

Carbon dioxide requires double bonds between carbon and each oxygen so that carbon achieves an octet with four bonding pairs total and each oxygen also completes its octet with two bonding pairs and two lone pairs. "Single bonds" would leave carbon with only two bonding pairs, giving it just four electrons and violating the octet rule. Drawing correct Lewis structures often requires testing multiple bond orders to satisfy the octet rule for every atom.

Q31. Why does the ionic compound \(\text{MgCl}_2\) have a formula with two chlorine atoms for every magnesium atom?
A Magnesium loses two electrons and each chlorine gains only one, so two chlorines balance the charge
B Magnesium loses one electron and chlorine gains two electrons
C Magnesium and chlorine share electrons equally in a 1:2 ratio
D Chlorine has twice the atomic mass of magnesium

Magnesium forms a \(2+\) ion by losing two valence electrons, and since each chlorine atom can only accept one electron to form a \(1-\) ion, two chloride ions are needed to balance the overall charge to neutral. The option stating chlorine "gains two electrons" is incorrect because chlorine only needs one electron to complete its octet. Balancing total positive and negative charge to zero is the key principle for determining ionic compound formulas.

Q32. Which molecule would you expect to have a nonpolar covalent bond based on electronegativity?
A \(\text{Cl}_2\)
B \(\text{HCl}\)
C \(\text{NaCl}\)
D \(\text{H}_2\text{O}\)

In \(\text{Cl}_2\), both atoms are identical chlorine atoms with the exact same electronegativity, so the bonding electrons are shared perfectly equally, making the bond nonpolar covalent. The option "\(\text{NaCl}\)" instead has a very large electronegativity difference between sodium and chlorine, making it ionic rather than covalent. Bonds between two atoms of the same element are always nonpolar covalent because there is zero electronegativity difference.

Q33. What is the correct total number of valence electrons to use when drawing the Lewis structure for \(\text{NH}_3\)?
A 8
B 5
C 3
D 10

Nitrogen contributes 5 valence electrons and each of the three hydrogen atoms contributes 1, giving a total of \(5 + 3(1) = 8\) valence electrons to distribute in the structure. The value "5" only accounts for nitrogen's electrons and ignores the hydrogens' contributions entirely. Summing valence electrons from all atoms in the formula is the essential first step before drawing any Lewis structure.

Q34. Why do noble gases rarely form chemical bonds?
A They already have a full valence shell of electrons, making them stable
B They have very high electronegativity values that repel other atoms
C They have too many protons to share electrons effectively
D They lose electrons too easily to form stable bonds

Noble gases already possess a complete octet (or duet for helium) of valence electrons, so they have no thermodynamic driving force to gain, lose, or share additional electrons through bonding. The claim that they have "very high electronegativity values that repel other atoms" is inaccurate because noble gases typically do not have meaningfully assigned electronegativity values due to their lack of bonding tendency. The octet rule explains both why most atoms bond to achieve noble gas configurations and why noble gases themselves remain largely unreactive.

Q35. What best explains why the bond in \(\text{F}_2\) is classified as nonpolar covalent despite fluorine being highly electronegative?
A Both atoms are identical, so the electronegativity difference between them is zero
B Fluorine atoms do not share electrons at all in this molecule
C Fluorine's high electronegativity always creates polar bonds regardless of the partner atom
D The bond in \(\text{F}_2\) is actually ionic, not covalent

Bond polarity depends on the difference in electronegativity between two bonded atoms, and since both atoms in \(\text{F}_2\) are fluorine, that difference is exactly zero, making the shared electrons distributed equally. The statement that fluorine "always creates polar bonds regardless of the partner atom" is false because polarity is a relative comparison, not an absolute property of one element alone. This example shows that even highly electronegative elements form nonpolar bonds when bonded to an identical atom.

Q36. Which statement correctly compares the strength of a double bond to a single bond between the same two atoms?
A A double bond is generally stronger and shorter than a single bond
B A double bond is generally weaker and longer than a single bond
C Double and single bonds have identical strength and length
D Bond strength cannot be compared between single and double bonds

A double bond involves two shared electron pairs instead of one, increasing the electron density between the nuclei, which pulls the atoms closer together and requires more energy to break, making it both shorter and stronger. The option claiming a double bond is "weaker and longer" reverses the correct trend seen experimentally in bond energy and bond length data. As bond order increases from single to double to triple, bond length decreases while bond strength increases.

Q37. In the compound \(\text{CaO}\), what is the primary reason the bond is classified as ionic?
A Calcium has low electronegativity and readily loses electrons while oxygen has high electronegativity and gains them
B Both calcium and oxygen have very similar electronegativities
C Calcium and oxygen share electrons equally in a covalent network
D Oxygen loses electrons to calcium, which is a nonmetal

Calcium is a metal with low electronegativity that easily loses its two valence electrons, while oxygen is a nonmetal with high electronegativity that readily accepts electrons, creating a large electronegativity difference characteristic of ionic bonding. The statement that oxygen "loses electrons to calcium, which is a nonmetal" is wrong on two counts, since oxygen gains rather than loses electrons and calcium is actually a metal. Identifying metal-nonmetal pairings with large electronegativity gaps is the standard method for predicting ionic bond formation.

Q38. What determines the number of bonds an atom typically forms according to the octet rule?
A The number of additional electrons needed to reach eight valence electrons
B The atomic mass of the element
C The number of protons in the nucleus
D The physical state of the element at room temperature

The octet rule states atoms tend to bond in ways that give them eight valence electrons, so the number of bonds an atom forms typically equals the number of electrons it needs to gain or share to complete that octet. "The atomic mass of the element" has no direct bearing on bonding behavior, since bonding depends on electron configuration, not mass. This principle explains why carbon, needing four more electrons, typically forms four bonds.

Q39. Which of the following correctly ranks bond polarity from least to most polar, based on typical electronegativity differences?
A \(\text{C-H} < \text{N-H} < \text{O-H}\)
B \(\text{O-H} < \text{N-H} < \text{C-H}\)
C \(\text{N-H} < \text{C-H} < \text{O-H}\)
D \(\text{O-H} < \text{C-H} < \text{N-H}\)

As electronegativity increases from carbon to nitrogen to oxygen, the electronegativity difference with hydrogen also increases, making the bonds progressively more polar in the order \(\text{C-H} < \text{N-H} < \text{O-H}\). The option placing \(\text{O-H}\) as least polar contradicts oxygen's high electronegativity, which creates the largest difference with hydrogen among these three bonds. Comparing electronegativity values along a row of the periodic table allows students to rank bond polarities without memorizing exact numbers.

Q40. Why is it necessary to place a formal negative charge on an atom in certain Lewis structures, such as in the polyatomic ion \(\text{NO}_3^-\)?
A The overall structure must account for the extra electron indicated by the ion's negative charge
B All Lewis structures require a formal charge regardless of the species
C Formal charges only apply to neutral molecules, not ions
D Negative charges are placed randomly to balance the drawing

Since \(\text{NO}_3^-\) carries an overall \(1-\) charge, one extra electron beyond the neutral atoms' valence electrons must be included in the total electron count and reflected somewhere in the structure through formal charge assignment. The claim that formal charges "only apply to neutral molecules, not ions" is incorrect, since formal charge calculations are especially important for accurately representing charged polyatomic ions. Tracking total charge, including extra or missing electrons, ensures the Lewis structure correctly represents the species being drawn.

Q41. Which best explains why potassium (K) is more reactive than sodium (Na) when forming ionic bonds?
A Potassium's valence electron is farther from the nucleus and more easily removed
B Potassium has a higher electronegativity than sodium
C Potassium has more valence electrons than sodium
D Potassium forms covalent bonds instead of ionic bonds

Potassium's valence electron occupies a higher principal energy level farther from the nucleus with more electron shielding, so it experiences a weaker attractive pull and is removed more easily than sodium's valence electron, making potassium more reactive in forming cations. The claim that potassium "has a higher electronegativity than sodium" is false, since electronegativity actually decreases going down a group. Reactivity trends among alkali metals in ionic bond formation are explained by decreasing ionization energy down a group.

Q42. What is the best explanation for why \(\text{SF}_6\) is a valid Lewis structure despite sulfur having twelve electrons around it, seemingly violating the octet rule?
A Sulfur is in period 3 or higher and can use its available d-orbitals to accommodate an expanded octet
B All elements can violate the octet rule without any exceptions or conditions
C Sulfur actually only has eight electrons around it in \(\text{SF}_6\)
D Fluorine atoms donate their electrons entirely to sulfur, forming an ionic compound

Elements in period 3 and beyond, like sulfur, have accessible d-orbitals that allow them to expand beyond the traditional octet and accommodate more than eight electrons, as seen with the six bonding pairs in \(\text{SF}_6\). The statement that "all elements can violate the octet rule without any conditions" is inaccurate, since this expanded octet exception is generally restricted to period 3 and higher elements, not elements like carbon or nitrogen in period 2. Recognizing expanded octets as a legitimate exception, tied to orbital availability, prevents students from forcing incorrect octet-only structures for larger central atoms.

Q43. A student proposes a Lewis structure for \(\text{SO}_2\) with two single bonds between sulfur and each oxygen, giving sulfur only six electrons. Why is this structure incorrect, and what is the fix?
A Sulfur needs a lone pair and one double bond to complete its octet while giving both oxygens a full octet
B The structure is correct as drawn since sulfur can have fewer than eight electrons
C Oxygen should form a triple bond to sulfur instead of single bonds
D The molecule should be redrawn as an ionic compound instead of covalent

To satisfy the octet rule, sulfur in \(\text{SO}_2\) needs one double bond and one single bond along with a lone pair, distributing electrons so both sulfur and each oxygen end up with eight electrons around them through resonance-averaged bonding. The claim that sulfur "can have fewer than eight electrons" as a stable, preferred structure is generally disfavored since incomplete octets are less stable than resonance structures that satisfy the octet rule for all atoms when possible. Testing initial single-bond-only structures against total electron count and octet completion often reveals the need for double or triple bonds.

Q44. Why does \(\text{CO}_2\) have polar bonds but is classified as a nonpolar molecule overall?
A The two C=O bond dipoles point in opposite directions and cancel due to the molecule's linear symmetry
B Carbon and oxygen have identical electronegativities, so no bond dipole exists
C \(\text{CO}_2\) is actually a polar molecule, not nonpolar
D The bonds in \(\text{CO}_2\) are purely ionic, so molecular polarity does not apply

Each C=O bond is individually polar due to oxygen's higher electronegativity, but because \(\text{CO}_2\) is linear with the two identical bond dipoles pointing in exactly opposite directions, the vector sum of the dipoles cancels to zero, making the overall molecule nonpolar. The claim that carbon and oxygen "have identical electronegativities" is false since oxygen is considerably more electronegative than carbon, which is exactly why each individual bond is polar. Molecular polarity depends on both bond polarity and molecular geometry together, not on bond polarity alone.

Q45. Which comparison best explains why \(\text{MgO}\) has a much higher melting point than \(\text{NaCl}\)?
A \(\text{MgO}\) ions have greater charges (\(2+\) and \(2-\)), producing stronger electrostatic attraction than the \(1+\) and \(1-\) charges in \(\text{NaCl}\)
B \(\text{MgO}\) has covalent bonds while \(\text{NaCl}\) has ionic bonds
C \(\text{NaCl}\) ions are smaller than \(\text{MgO}\) ions, weakening its lattice
D Melting point is unrelated to ionic charge in these compounds

According to Coulomb's law, electrostatic attraction between ions increases with the product of their charges, so the \(2+\) and \(2-\) charges in \(\text{MgO}\) create a much stronger lattice energy and higher melting point than the singly charged ions in \(\text{NaCl}\). The claim that "\(\text{MgO}\) has covalent bonds while \(\text{NaCl}\) has ionic bonds" is incorrect since both compounds are ionic, formed between metals and nonmetals with large electronegativity differences. Comparing ionic charge magnitude is one of the most reliable ways to predict relative lattice energy and melting point trends among ionic compounds.

Q46. Why does the molecule \(\text{NH}_3\) have a higher boiling point than \(\text{PH}_3\), despite both having a similar trigonal pyramidal shape?
A Nitrogen's high electronegativity and small size allow \(\text{NH}_3\) to form strong hydrogen bonds, unlike \(\text{PH}_3\)
B Phosphorus is more electronegative than nitrogen, giving \(\text{PH}_3\) weaker intermolecular forces
C \(\text{NH}_3\) has ionic bonds while \(\text{PH}_3\) has covalent bonds
D Molecular shape alone determines boiling point, and shape differences explain this trend

Nitrogen's small atomic radius and high electronegativity allow \(\text{NH}_3\) molecules to form strong hydrogen bonds between molecules, a much stronger intermolecular force than the weaker dipole-dipole and dispersion forces present in \(\text{PH}_3\), resulting in a higher boiling point for ammonia. The claim that "phosphorus is more electronegative than nitrogen" is false, since nitrogen is actually more electronegative than phosphorus on the periodic table. Recognizing hydrogen bonding as a uniquely strong intermolecular force involving N-H, O-H, or F-H bonds is essential for predicting relative boiling points.

Q47. A compound has the formula \(\text{AlCl}_3\) but exhibits some covalent character rather than being purely ionic. What best explains this observation?
A Aluminum's relatively small size and high charge density polarize the electron cloud of chloride ions, pulling shared electron density toward aluminum
B Aluminum and chlorine have identical electronegativities, making the bond purely covalent
C \(\text{AlCl}_3\) is actually a metallic compound, not ionic or covalent
D All metal-nonmetal compounds are purely ionic with no exceptions

Aluminum's small ionic radius combined with its \(3+\) charge creates a high charge density that strongly polarizes the electron cloud of the neighboring chloride ions, distorting the electron distribution enough to introduce significant covalent character into the bond, a trend described by Fajans' rules. The statement that "all metal-nonmetal compounds are purely ionic with no exceptions" oversimplifies bonding, since real bonds exist on a continuum and highly charged, small metal cations like \(\text{Al}^{3+}\) often show partial covalent behavior. This illustrates that the ionic-covalent classification is a spectrum rather than a strict binary, especially for cations with high charge density.

Q48. Why do resonance structures for the carbonate ion \(\text{CO}_3^{2-}\) better represent its actual bonding than any single Lewis structure?
A The true structure is a hybrid average of all resonance forms, with bond lengths intermediate between single and double bonds
B Only one of the resonance structures is actually correct, and the others should be discarded
C Resonance structures indicate the molecule rapidly switches between forms, changing shape every instant
D Carbonate does not actually have resonance and is fully represented by a single Lewis structure

Experimental data shows all three carbon-oxygen bonds in \(\text{CO}_3^{2-}\) have equal, intermediate bond lengths, which is best explained by treating the actual structure as a resonance hybrid, an average of the contributing Lewis structures rather than any single one being literally correct. The claim that the molecule "rapidly switches between forms, changing shape every instant" misrepresents resonance, since the real structure is a single, static blended hybrid rather than an oscillation between distinct structures. Resonance structures are a modeling tool to represent delocalized electron density that no single classical Lewis structure can fully capture.

Q49. Why does hydrogen fluoride (\(\text{HF}\)) have unusually strong intermolecular attractions compared to other hydrogen halides like \(\text{HCl}\)?
A Fluorine's extremely high electronegativity and small size create a highly polarized H-F bond that enables strong hydrogen bonding
B Fluorine forms ionic bonds with hydrogen, unlike chlorine which forms covalent bonds
C \(\text{HF}\) molecules are larger than \(\text{HCl}\) molecules, increasing dispersion forces
D Hydrogen bonding does not occur between \(\text{HF}\) molecules

Fluorine's very high electronegativity combined with its small atomic size creates a highly polarized H-F bond, concentrating a strong partial negative charge on a small fluorine atom that allows exceptionally strong hydrogen bonding between HF molecules. The claim that fluorine "forms ionic bonds with hydrogen, unlike chlorine" is incorrect, since both H-F and H-Cl bonds are polar covalent, differing only in the degree of polarity. Among hydrogen halides, only \(\text{HF}\) exhibits significant hydrogen bonding because fluorine meets the small-size, high-electronegativity criteria required for this strong intermolecular force.

Q50. Which explanation best accounts for why graphite conducts electricity while diamond, another form of pure carbon, does not?
A Graphite has delocalized pi electrons free to move within its layered structure, while diamond's electrons are all localized in fixed covalent bonds
B Graphite is an ionic compound while diamond is purely covalent
C Diamond has more valence electrons per carbon atom than graphite
D Graphite contains metallic bonds throughout its entire structure

In graphite, each carbon forms three sigma bonds within a planar layer, leaving one electron delocalized in a pi system that can move freely across the layer, giving graphite electrical conductivity, while diamond's four localized sigma bonds per carbon leave no free electrons for conduction. The claim that graphite "is an ionic compound while diamond is purely covalent" is false, since both allotropes are entirely composed of covalently bonded carbon atoms, just arranged differently. Comparing carbon allotropes demonstrates how bonding geometry and electron delocalization, not just bond type, determine macroscopic properties like conductivity.

Q51. Why is the bond length in the nitrogen molecule \(\text{N}_2\) shorter than the bond length in the oxygen molecule \(\text{O}_2\)?
A Nitrogen forms a triple bond with three shared electron pairs, while oxygen forms only a double bond with two shared pairs
B Oxygen atoms are smaller than nitrogen atoms, making the \(\text{O}_2\) bond inherently shorter
C Nitrogen has a lower electronegativity than oxygen, which shortens its bond
D Bond length depends only on atomic mass, and oxygen is heavier than nitrogen

Nitrogen achieves a stable octet by forming a triple bond, sharing three electron pairs between the two atoms, which pulls the nuclei closer together than the double bond in \(\text{O}_2\), which only shares two electron pairs. The claim that "oxygen atoms are smaller than nitrogen atoms" is not the primary driver here and is actually inaccurate, since atomic radius alone does not explain this trend as strongly as bond order does. Higher bond order consistently correlates with shorter, stronger bonds when comparing bonds of the same atom pair type.

Q52. A student claims that because \(\text{BeCl}_2\) has a metal bonded to a nonmetal, it must be purely ionic. What is the flaw in this reasoning?
A Beryllium's very small size and high charge density give it significant covalent character despite being a metal
B All metal-nonmetal bonds are always purely ionic with no exceptions
C Chlorine's electronegativity is too low to attract electrons from beryllium
D \(\text{BeCl}_2\) is actually a purely covalent compound with no ionic character at all

Beryllium is an unusually small metal cation with high charge density that strongly polarizes the electron cloud of chloride ions, pulling electron density back toward itself and giving the bond substantial covalent character rather than being purely ionic. The blanket assumption that "all metal-nonmetal bonds are always purely ionic" ignores well-documented exceptions like beryllium and aluminum compounds, where small, highly charged cations distort anions significantly. Bond character exists on a continuum influenced by both electronegativity difference and ionic size or charge, not determined by metal-nonmetal classification alone.

Q53. Why can the Lewis structure of ozone (\(\text{O}_3\)) not be fully represented by a single structure with one double bond and one single bond?
A Experimental bond lengths show both oxygen-oxygen bonds are equal and intermediate, requiring a resonance hybrid description
B Ozone only has single bonds between all oxygen atoms
C A single structure with two double bonds satisfies the octet rule perfectly for ozone
D Ozone does not follow the octet rule at all, making Lewis structures inapplicable

Measured bond lengths in ozone show both O-O bonds are identical and intermediate between typical single and double bond lengths, which cannot be captured by any single Lewis structure with one localized double bond, requiring instead an average of two resonance structures. The option stating ozone "only has single bonds between all oxygen atoms" fails to satisfy the octet rule for the central oxygen atom, which needs one double bond contribution to complete its octet. Resonance is invoked precisely when experimental structural data conflicts with what any single Lewis structure predicts.

Q54. Two compounds, \(\text{KF}\) and \(\text{CaO}\), are compared for lattice energy. Which factor most strongly explains why \(\text{CaO}\) has the higher lattice energy?
A \(\text{CaO}\) has ions with greater charge magnitude (\(2+\)/\(2-\)) compared to the singly charged ions in \(\text{KF}\)
B \(\text{KF}\) ions are larger, which increases its lattice energy above that of \(\text{CaO}\)
C Potassium and calcium have identical charges, so ionic radius is the only relevant factor
D Lattice energy depends solely on the type of nonmetal present, not the metal

Lattice energy scales with the product of ionic charges according to Coulomb's law, so the doubly charged \(\text{Ca}^{2+}\) and \(\text{O}^{2-}\) ions in \(\text{CaO}\) create substantially stronger electrostatic attraction and higher lattice energy than the singly charged \(\text{K}^+\) and \(\text{F}^-\) ions in \(\text{KF}\). The claim that "potassium and calcium have identical charges" is factually wrong, since potassium forms a \(1+\) ion while calcium forms a \(2+\) ion, a key difference driving the lattice energy comparison. When comparing lattice energies, ionic charge typically has a greater impact than ionic size differences of similar magnitude.

Q55. How many valence electrons does chlorine have?
A 7
B 5
C 8
D 1

Chlorine is located in Group 17 (VIIA) of the periodic table, meaning it has 7 valence electrons in its outer energy level. The option "8" would represent a complete octet, which chlorine reaches only after gaining one more electron through bonding. Group number provides a fast way to determine the valence electron count needed to predict bonding behavior.

Q56. What is formed when two nonmetal atoms bond by sharing electrons?
A A covalent bond
B An ionic bond
C A metallic bond
D A hydrogen atom

When two nonmetal atoms come together, both have relatively high electronegativity and neither easily gives up electrons, so they share electron pairs to complete their octets, forming a covalent bond. "An ionic bond" instead requires one atom to transfer electrons completely to another, which typically occurs between a metal and a nonmetal. Nonmetal-nonmetal combinations are a reliable indicator that a covalent bond, rather than an ionic one, will form.

Q57. Which element has the highest electronegativity on the periodic table?
A Fluorine
B Oxygen
C Chlorine
D Nitrogen

Fluorine has the highest electronegativity of any element, roughly \(3.98\) on the Pauling scale, due to its small atomic radius and strong effective nuclear charge on valence electrons. "Oxygen" is highly electronegative as well but ranks below fluorine on the standard electronegativity scale. Fluorine's position in the top-right corner of the periodic table (excluding noble gases) makes it the benchmark for maximum electronegativity.

Q58. What is a lone pair of electrons?
A A pair of valence electrons on an atom not involved in bonding
B A pair of electrons shared between two bonded atoms
C A single unpaired electron in an orbital
D A pair of electrons transferred during ionic bond formation

A lone pair consists of two valence electrons that belong to a single atom and are not shared with another atom in a bond, often shown as dots in Lewis structures. The choice "A pair of electrons shared between two bonded atoms" instead describes a bonding pair, which forms a covalent bond rather than remaining unshared. Distinguishing lone pairs from bonding pairs is critical for correctly predicting molecular geometry using VSEPR theory.

Q59. Which of these compounds is held together primarily by ionic bonds?
A \(\text{KBr}\)
B \(\text{CH}_4\)
C \(\text{CO}_2\)
D \(\text{N}_2\)

Potassium bromide (\(\text{KBr}\)) forms between a metal, potassium, and a nonmetal, bromine, resulting in a large electronegativity difference and the transfer of electrons that characterizes an ionic bond. "\(\text{CH}_4\)" is composed entirely of nonmetals, carbon and hydrogen, which share electrons covalently rather than transferring them. Recognizing metal-nonmetal combinations as the classic signature of ionic bonding helps quickly classify unfamiliar compounds.

Q60. When drawing a Lewis structure, which atom is typically placed in the center of the molecule?
A The atom with the lowest electronegativity (excluding hydrogen)
B The atom with the highest electronegativity
C Hydrogen, since it usually forms the most bonds
D The atom with the largest atomic mass

The central atom in a Lewis structure is generally the one with the lowest electronegativity, excluding hydrogen, because less electronegative atoms are typically better at forming multiple bonds to surrounding atoms. The option "Hydrogen, since it usually forms the most bonds" is incorrect because hydrogen can only form one bond, since it needs just two electrons total to satisfy its duet rule. Placing hydrogen only as a terminal (outer) atom and choosing the least electronegative other atom as central is a standard strategy for building Lewis structures.

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Quick summary

This unit covers ionic bonds, covalent bonds, Lewis structures and electronegativity — essential concepts for Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.

Key concepts
  • Ionic bonds
  • Covalent bonds
  • Lewis structures
  • Electronegativity
What you need to know

Key Concepts Breakdown

1 Ionic Bonds

Ionic bonds form between a metal and a nonmetal through the transfer of electrons, creating oppositely charged ions that attract each other. Students must know that metals lose electrons to form cations and nonmetals gain electrons to form anions. The resulting compound is electrically neutral overall.

Key Points

  • Metal loses electrons → cation (positive charge); nonmetal gains electrons → anion (negative charge)
  • Ionic compounds form crystal lattice structures and have high melting points
  • Ionic compounds conduct electricity when dissolved in water or melted
  • The charge of each ion is determined by how many electrons are lost or gained to reach a full outer shell
Example

Predict the formula for the ionic compound formed between calcium (Ca) and chlorine (Cl).

Explanation

Calcium is in Group 2, so it loses 2 electrons to form Ca²⁺. Chlorine is in Group 17, so it gains 1 electron to form Cl⁻. To balance the charges, you need 1 Ca²⁺ and 2 Cl⁻, giving the formula CaCl₂.

2 Covalent Bonds

Covalent bonds form between two nonmetals through the sharing of electron pairs. Students must distinguish between single (2 electrons shared), double (4 electrons shared), and triple bonds (6 electrons shared), and understand that more shared pairs create shorter, stronger bonds.

Key Points

  • Nonmetal + nonmetal → covalent bond (electron sharing, not transfer)
  • Single bond < double bond < triple bond in terms of bond strength and shortness
  • Molecules formed by covalent bonds generally have lower melting points than ionic compounds
  • Polar covalent bonds occur when electrons are shared unequally due to differing electronegativities
Example

Classify the bond in O₂ and state how many electrons are shared.

Explanation

Oxygen and oxygen are both nonmetals, so they form a covalent bond. Each oxygen needs 2 more electrons to complete its octet, so they share 2 pairs of electrons, forming a double bond with 4 electrons total shared between them.

3 Lewis Structures

Lewis structures are diagrams that show how valence electrons are arranged around atoms in a molecule, including bonding pairs and lone pairs. Students must be able to draw them correctly by counting total valence electrons, forming bonds, and satisfying the octet rule for each atom (with hydrogen limited to 2 electrons).

Key Points

  • Count total valence electrons: use the periodic group number for each atom, adjust for charge if ion
  • Hydrogen gets 2 electrons (duet rule); most other atoms follow the octet rule (8 electrons)
  • Place lone pairs on outer atoms first, then on the central atom
  • If the central atom lacks an octet, convert lone pairs to double or triple bonds
Example

Draw the Lewis structure for CO₂.

Explanation

Carbon has 4 valence electrons and each oxygen has 6, giving 4 + 6 + 6 = 16 total valence electrons. Carbon is the central atom; placing single bonds to each oxygen uses 4 electrons, leaving 12 for lone pairs. Filling oxygen octets uses all 12, but carbon only has 4 electrons — so each lone pair on oxygen is converted to a shared pair, forming two double bonds (O=C=O), satisfying all octets with 16 electrons used.

4 Electronegativity

Electronegativity is a measure of how strongly an atom attracts shared electrons in a bond. Students must know the general trend (increases across a period, decreases down a group) and use electronegativity differences to classify bonds as nonpolar covalent, polar covalent, or ionic.

Key Points

  • Electronegativity increases across a period (left → right) and decreases down a group (top → bottom); fluorine is highest
  • ΔEN = 0–0.4: nonpolar covalent bond; ΔEN = 0.5–1.7: polar covalent bond; ΔEN > 1.7: ionic bond
  • In a polar covalent bond, the more electronegative atom carries a partial negative charge (δ⁻)
  • Bond polarity affects molecular polarity, which affects physical properties like boiling point and solubility
Example

Classify the bond between hydrogen (EN = 2.1) and fluorine (EN = 4.0). Which atom is δ⁻?

Explanation

The electronegativity difference is 4.0 − 2.1 = 1.9, which is greater than 1.7, so this bond is ionic in character — though HF is typically classified as polar covalent in introductory courses because it is a molecule. Fluorine has the higher electronegativity, so it pulls the shared electrons closer to itself and carries the partial negative charge (δ⁻), while hydrogen is δ⁺.

FAQ

Questions, answered.

What is Chemical Bonding?

Chemical Bonding is Unit 3 of Chemistry, covering ionic bonds, covalent bonds, Lewis structures and electronegativity.

How to study for Chemistry Unit 3?

Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.

How many questions are in this unit?

This unit has 60 review questions, each with a written explanation, playable across 5 different game modes or readable in plain-text mode.